Question

For the titration of 45.0 mL of 0.350 M NH3 with 0.500 M HCl at 25 °C, determine the relative pH at each of these points.

Answer 1

(a) before the addition of any HCl:  pH>7
(b) after 18.0 mL of HCl has been added : pH<7 
(c) after 33.0 mL of HCl has been added : pH<7

Answer 2

The Kb for ammonia is 1.8 X 10⁻⁵

The ammonia will dissociate as

                  NH₃ + H₂O ---> NH₄⁺ + OH⁻

initial            0.350                0           0

Change           -x                     +x        +x

Equilibrium    0.35-x                 x        x

$Kb=\frac{[NH_{4}^{+}[OH^{-} ]}{[NH_{3}]}$

Putting values

$Kb=\frac{[x][x]}{[0.35-x]}=\frac{x^{2} }{0.35-x}$

We may ignore x in denominator as Kb is very low

therefore

x² = 1.8 X 10⁻⁵  X 0.35  = 0.63 X 10⁻⁵

x = 2.51 X 10⁻³ M

[OH⁻] = 2.51 X 10⁻³

pOH = -log[OH⁻] = 2.6

pH = 14- 2.6 = 11.4

b) after addition of 18 mL of HCl

The moles of acid added = molarity X volume = 0.5 x 18 = 9 mmol

the moles of base present = molarity X volume = 0.35 X 45 = 15.75 mmol

so moles of base left = 6.75 mmol

moles of salt formed = 9 mmol

This will results in formation of buffer

the pH of buffer is calculated using

pOH = pKb + log [salt] / [base]

pOH = 4.74 + log [9/6.75] = 4.86

pH = 9.14

c) on addition of 33 mL of HCl

The moles of acid added = molarity X volume = 0.5 x 33 = 16.5 mmol

the moles of base present = molarity X volume = 0.35 X 45 = 15.75 mmol

Hence the acid will completely neutralize the base

The acid left = 16.5 - 15.75 = 0.75 mmol

total volume = 33 + 45 = 75mL

[HCl] = 0.75 / 75 = 0.01 M

pH = -log[0.01] = 2